In the following reaction Cr_2O_7^{2-} + 14H^ | vedclass

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(A) To determine which element is reduced,we calculate the oxidation numbers of the elements in the reaction.In $Cr_2O_7^{2-}$,the oxidation state of

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$Cr$

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(A) To determine which element is reduced,we calculate the oxidation numbers of the elements in the reaction.In $Cr_2O_7^{2-}$,the oxidation state of $Cr$ is $+6$.In $2Cr^{3+}$,the oxidation state of $Cr$ is $+3$.Since the oxidation state of $Cr$ decreases from $+6$ to $+3$,it undergoes reduction.Therefore,$Cr$ is the element that is reduced.

In the following reaction $Cr_2O_7^{2-} + 14H^{+} + 6I^{-} \to 2Cr^{3+} + 7H_2O + 3I_2$,which element is reduced?

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